Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
Lesson 6: Probability Rules
Student Outcomes
Students use the complement rule to calculate the probability of the complement of an event and the
multiplication rule for independent events to calculate the probability of the intersection of two independent
events.
Students recognize that two events 𝐴 and 𝐵 are independent if and only if 𝑃(𝐴 and 𝐵) = 𝑃(𝐴)𝑃(𝐵) and
interpret independence of two events 𝐴 and 𝐵 as meaning that the conditional probability of 𝐴 given 𝐵 is
equal to 𝑃(𝐴).
Students use the formula for conditional probability to calculate conditional probabilities and interpret
probabilities in context.
Lesson Notes
This lesson introduces the formulas for calculating the probability of the complement of an event, the probability of an
intersection when events are independent, and conditional probabilities. Because these concepts have already been
presented using the more intuitive approach of earlier lessons, students should readily understand why the formulas are
true.
Classwork
Opening (4 minutes)
Begin this lesson by asking students the following questions:
There are 300 students at a certain school. All students indicated they were
either right-handed or left-handed but not both. Fifty of the students are lefthanded. How many students are right-handed? What is the probability of
randomly selecting a right-handed student at this school? How did you
determine this?
The United States Census Bureau indicates there are 19,378,102 people in the
State of New York. If 4,324,929 are under the age of 18, how many are 18 and
older? What is the probability of randomly selecting a person from New York
who is 18 or older? How did you determine this?
It is estimated that approximately 20% of the people in the United States have
asthma and severe allergies. What is the probability that a randomly selected
person does not have asthma or a severe allergy? How did you determine this?
Lesson 6:
Date:
Scaffolding:
For students that are struggling
with this concept, consider
displaying a visual of the
numerical representation of
each problem (e.g., 300 – 50,
19,378,102 – 4,324,929,
100% – 20%) and asking
them, “What do all of these
expressions have in common?”
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
Discuss with students how these examples, or similar examples, involve finding the
probability of the complement of a given event. Work with students in deriving the
probabilities by thinking of the probability of people who are not in the given
probability and connecting that probability to the fact that the two probabilities (or
the probability of the given event and the probability of the complement) add up to
1.00 or 100%. Although students have previously worked with this idea, this lesson
will formalize their understanding of complement and the probability of
complementary events.
Use this as an opportunity to have students look for a general method or rule for
MP.8 determining the probability of the complement of an event. Encourage students to
try to write a rule and ask them to share it with their neighbor.
Based on the three examples we looked at, what is a general rule we could
write to determine the probability of the complement of an event?
Example 1 (3 minutes): The Complement Rule
At this point, students have informally used the complement rule, but in this example
it is given as a formula. The example serves as a quick illustration of the rule. Read
through the example as a class, and then give students a moment to calculate the
probability presented at the end of the example.
Scaffolding:
Clarify that the term
“complement” is not the same
as “compliment.” Point out
the difference in spelling
between the terms, as well
their meanings.
The complement of any event
is the event that does not
occur.
A compliment is an expression
of respect or admiration.
Consider using visual displays
and repeated choral readings
to reinforce the mathematical
meaning of the word.
A Frayer model may also be
used.
Example 1: The Complement Rule
In previous lessons you have seen that to calculate the probability that an event does not
happen you can subtract the probability of the event from 1. If the event is denoted by 𝑨,
then this rule can be written:
𝑷(not 𝑨) = 𝟏 − 𝑷(𝑨).
For example, suppose that the probability that a particular flight is on time is 𝟎. 𝟕𝟖. What is the probability that the flight
is not on time?
𝑷(not on time) = 𝟏 − 𝟎. 𝟕𝟖 = 𝟎. 𝟐𝟐
Scaffolding:
Example 2 (6 minutes): Formula for Conditional Probability
The purpose of this example is to introduce the formula for conditional probability. A
conditional probability is first calculated using a hypothetical 1000 table (as in
previous lessons), and then the formula is shown to produce the same result.
When working through part (c) with the class, it would be helpful to illustrate the
division with the aid of a Venn diagram so that students get a visual idea of what is
being divided by what. (The probability of the intersection is being divided by the
probability of 𝐵.)
Consider asking students to try solving the entire problem or parts and then discussing
the results as a class.
Lesson 6:
Date:
In addition to using Venn
diagrams, if students have
trouble identifying the
intersection of two events,
point out real-world examples
like intersecting streets to
connect students to the
concept.
Students may confuse the term
conditional with conditioner or
“condition.” Point out the
difference in spelling and
meaning.
Again the use of visuals and
repeated choral readings will
help reinforce these words.
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
Additionally, point out that
0.38
0.43
is exactly the same thing as
380
430
(the numerator and the denominator have been divided
by 1,000). By seeing this students should see why the conditional probability formula is valid. (This is the main point of
this example.)
Example 2: Formula for Conditional Probability
When a room is randomly selected in a downtown hotel, the probability that the room has a king-sized bed is 𝟎. 𝟔𝟐, the
probability that the room has a view of the town square is 𝟎. 𝟒𝟑, and the probability that it has a king-sized bed and a
view of the town square is 𝟎. 𝟑𝟖. Let 𝑨 be the event that the room has a king-sized bed, and let 𝑩 be the event that the
room has a view of the town square.
a.
What is the meaning of “𝑷(𝑨 given 𝑩)” in this context?
𝑷(𝑨 given 𝑩) is the probability that a room known to have a view of the town square also has a king-sized bed.
b.
Use a hypothetical 1000 table to calculate 𝑷(𝑨 given 𝑩).
𝑨
(room has a king-sized bed)
Not 𝑨
(room does not have kingsized bed)
Total
𝟑𝟖𝟎
𝟓𝟎
𝟒𝟑𝟎
𝟐𝟒𝟎
𝟑𝟑𝟎
𝟓𝟕𝟎
𝟔𝟐𝟎
𝟑𝟖𝟎
𝟏𝟎𝟎𝟎
𝑩
(room has a view of the
town square)
Not 𝑩
(room does not have view
of the town square)
Total
𝑷(𝑨 given 𝑩) =
c.
𝟑𝟖𝟎
= 𝟎. 𝟖𝟖𝟒
𝟒𝟑𝟎
There is also a formula for calculating a conditional probability. The formula for conditional probability is:
𝑷(𝑨 given 𝑩) =
𝑷(𝑨 and 𝑩)
.
𝑷(𝑩)
Use this formula to calculate 𝑷(𝑨 given 𝑩), where the events 𝑨 and 𝑩 are as defined in this example.
𝑷(𝑨 given 𝑩) =
d.
𝑷(𝑨 and 𝑩) 𝟎. 𝟑𝟖
=
= 𝟎. 𝟖𝟖𝟒
𝑷(𝑩)
𝟎. 𝟒𝟑
How does the probability you calculated using the formula compare to the probability you calculated using the
hypothetical 1000 table?
The probabilities are the same.
Exercise 1 (15 minutes)
This exercise provides practice using the conditional probability formula. Additionally, in part (e) students are asked to
compare a conditional and an unconditional probability and to provide an interpretation. In part (f) students are asked
to recall the definition of independence in terms of equality of the conditional and unconditional probabilities.
Let students work with a partner and then confirm answers as a class, spending about 10 minutes total on the exercise.
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Lesson 6
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M4
ALGEBRA II
After confirming the answer to part (f), present the multiplication rule for independent events (approximately 5
minutes). Use this as an opportunity to informally assess student understanding of the lesson’s outcomes.
Exercise 1
A credit card company states that 𝟒𝟐% of its customers are classified as long-term cardholders, 𝟑𝟓% pay their bills in full
each month, and 𝟐𝟑% are long-term cardholders who also pay their bills in full each month. Let the event that a
randomly selected customer is a long-term cardholder be 𝑳, and the event that a randomly selected customer pays his or
her bill in full each month be 𝑭.
a.
What are the values of 𝑷(𝑳), 𝑷(𝑭), and 𝑷(𝑳 and 𝑭)?
𝑷(𝑳) = 𝟎. 𝟒𝟐, 𝑷(𝑭) = 𝟎. 𝟑𝟓, 𝑷(𝑳 and 𝑭) = 𝟎. 𝟐𝟑
b.
Draw a Venn diagram, and label it with the probabilities from part (a).
c.
Use the conditional probability formula to calculate 𝑷(𝑳 given 𝑭). (Round your answer to the nearest
thousandth.)
𝑷(𝑳 given 𝑭) =
d.
Use the conditional probability formula to calculate 𝑷(𝑭 given 𝑳). (Round your answer to the nearest
thousandth.)
𝑷(𝑭 given 𝑳) =
e.
𝑷(𝑳 and 𝑭) 𝟎. 𝟐𝟑
=
= 𝟎. 𝟔𝟓𝟕
𝑷(𝑭)
𝟎. 𝟑𝟓
𝑷(𝑭 and 𝑳) 𝟎. 𝟐𝟑
=
= 𝟎. 𝟓𝟒𝟖
𝑷(𝑳)
𝟎. 𝟒𝟐
Which is greater, 𝑷(𝑭 given 𝑳) or 𝑷(𝑭)? Explain why this is relevant.
𝑷(𝑭 given 𝑳) = 𝟎. 𝟓𝟒𝟖 and 𝑷(𝑭) = 𝟎. 𝟑𝟓; therefore, 𝑷(𝑭 given 𝑳) is larger than 𝑷(𝑭). This tells us that longterm cardholders are more likely to pay their bill in full each month than customers in general.
f.
Remember that two events A and B are said to be independent if 𝑷(𝑨 given 𝑩) = 𝑷(𝑨). Are the events
𝑭 and 𝑳 independent? Explain.
Note that there are several ways to answer this question. Here are three possibilities:
No, because 𝑷(𝑭 given 𝑳) ≠ 𝑷(𝑭).
No, because 𝑷(𝑳 given 𝑭) ≠ 𝑷(𝑳).
No, because 𝑷(𝑳 and 𝑭) ≠ 𝑷(𝑳)𝑷(𝑭).
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Lesson 6
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M4
ALGEBRA II
After confirming answers to part (f), introduce the multiplication rule for independent
events:
Scaffolding:
Events 𝐴 and 𝐵 are independent if and only if 𝑃(𝐴 and 𝐵) = 𝑃(𝐴)𝑃(𝐵).
This means that if we know the events 𝐴 and 𝐵 are independent, then we can conclude
that 𝑃(𝐴 and 𝐵) = 𝑃(𝐴)𝑃(𝐵), and also that if we know that 𝑃(𝐴 and 𝐵) = 𝑃(𝐴)𝑃(𝐵),
then we can conclude that the events 𝐴 and 𝐵 are independent. Consider asking students
to state the meaning of this rule in their own words. The easiest way to explain this is as
follows.
For students that may be
above grade level, consider
asking the following:
“Explain why this statement
makes sense using conditional
probabilities.”
(The table below offers a
sample explanation.)
𝑃(𝐴 given 𝐵) = 𝑃(𝐴)
If the events 𝐴 and 𝐵 are independent then we know:
Use the formula for conditional probability to replace 𝑃(𝐴 given 𝐵):
𝑃(𝐴 given 𝐵) =
𝑃(𝐴 and 𝐵)
𝑃(𝐵)
𝑃(𝐴 and 𝐵)
= 𝑃(𝐴)
𝑃(𝐵)
𝑃(𝐴 and 𝐵) = 𝑃(𝐴)𝑃(𝐵)
Now isolate 𝑃(𝐴 and 𝐵) and conclude that
Example 3 (5 minutes): Using the Multiplication Rule for Independent Events
This is an example of the use of the multiplication rule for independent events. There are two ways of telling whether
events are independent: Either it is obvious from the description of the problem (as in part (a)), or the question tells you
that the events are independent (as in part (b)). Consider asking students to attempt to solve independently or with a
neighbor, informally assessing understanding and offering guidance as necessary. When tackling part (a):
Explain to students that the result for the number cube cannot possibly affect the result for the coin, and so
the two events are independent.
Explain that the result of any one roll of the number cube cannot have an effect on the results of the other two
rolls.
Example 3: Using the Multiplication Rule for Independent Events
A number cube has faces numbered 𝟏 through 𝟔, and a coin has two sides, heads and tails.
The number cube will be rolled and the coin will be flipped. Find the probability that the cube shows a 𝟒, and the coin
lands heads. Because the events are independent, we can use the multiplication rules we just learned.
𝟏 𝟏
𝟏
( )( ) =
𝟔 𝟐
𝟏𝟐
If you toss the coin five times, what is the probability you will see a head on all five tosses?
(𝟎. 𝟓)(𝟎. 𝟓)(𝟎. 𝟓)(𝟎. 𝟓)(𝟎. 𝟓) = 𝟎. 𝟎𝟑𝟏𝟐𝟓
If you tossed the coin five times and got five heads, would you think that this coin is a fair coin? Why or why not?
Although getting five heads is possible (about 𝟑 percent of the time you would expect this), it is not likely, and, therefore,
you suspect that the coin is not fair.
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
If you roll the number cube three times, what is the probability that it will show 𝟒 on all three throws?
𝟏 𝟏 𝟏
𝟏
( )( )( ) =
≈ 𝟎. 𝟎𝟎𝟓
𝟔 𝟔 𝟔
𝟐𝟏𝟔
If you rolled the number cube three times and got a 𝟒 on all three rolls, would you think that this number cube is fair?
Why or why not?
The probability of getting a 𝟒 on all three rolls is very small. As a result, you suspect the number cube is not fair.
Suppose that the credit card company introduced in Exercise 1 states that when a customer is selected at random, the
probability that the customer pays his or her bill in full each month is 𝟎. 𝟑𝟓, the probability that the customer makes
regular online purchases is 𝟎. 𝟖𝟑, and these two events are independent. What is the probability that a randomly
selected customer pays his or her bill in full each month and makes regular online purchases?
(𝟎. 𝟑𝟓)(𝟎. 𝟖𝟑) = 𝟎. 𝟐𝟗𝟎𝟓
Exercise 2 (5 minutes)
This exercise provides practice with use of the multiplication rule for independent events. Students should think about
why the events given are independent (in this case, because it is obvious from the problem description). Particularly in
MP.4
part (c), students have an opportunity to interpret their results in the context of the question and to reflect on whether
the results make sense. Consider asking students to work independently on this exercise.
Exercise 2
A spinner has a pointer, and when the pointer is spun, the probability that it stops in the red section of the spinner is
𝟎. 𝟐𝟓.
a.
If the pointer is spun twice, what is the probability that it will stop in the red section on both occasions?
(𝟎. 𝟐𝟓)(𝟎. 𝟐𝟓) = 𝟎. 𝟎𝟔𝟐𝟓
b.
If the pointer is spun four times, what is the probability that it will stop in the red section on all four
occasions? (Round your answer to the nearest thousandth.)
(𝟎. 𝟐𝟓)(𝟎. 𝟐𝟓)(𝟎. 𝟐𝟓)(𝟎. 𝟐𝟓) = 𝟎. 𝟎𝟎𝟒
c.
If the pointer is spun five times, what is the probability that it never stops on red? (Round your answer to the
nearest thousandth.)
(𝟎. 𝟕𝟓)(𝟎. 𝟕𝟓)(𝟎. 𝟕𝟓)(𝟎. 𝟕𝟓)(𝟎. 𝟕𝟓) = 𝟎. 𝟐𝟑𝟕
Closing (2 minutes)
Ask students to summarize the main ideas of the lesson in writing or with a neighbor. Use this as an opportunity to
informally assess comprehension of the lesson. The Lesson Summary below offers some important ideas that should be
included.
Does it make sense to consider the two events in Example 3 part (b) to be independent? (Events are that
customer pays bill in full each month and makes regular online purchases.)
Yes, it seems feasible that long-term cardholders might be as likely to make regular online purchases as
customers in general.
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NYS COMMON CORE MATHEMATICS CURRICULUM
Lesson 6
M4
ALGEBRA II
Lesson Summary
For any event 𝑨, 𝑷(not 𝑨) = 𝟏 − 𝑷(𝑨).
For any two events 𝑨 and 𝑩, 𝑷(𝑨 given 𝑩) =
𝑷(𝑨 and 𝑩)
.
𝑷(𝑨)
Events 𝑨 and 𝑩 are independent if and only if 𝑷(𝑨 and 𝑩) = 𝑷(𝑨)𝑷(𝑩).
Exit Ticket (5 minutes)
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
Name
Date
Lesson 6: Probability Rules
Exit Ticket
1.
Of the light bulbs available at a store, 42% are fluorescent, 23% are labeled as “long life,” and 12% are fluorescent
and “long life.”
a.
b.
2.
A light bulb will be selected at random from the light bulbs at this store. Rounding your answer to the nearest
thousandth where necessary, find the probability that
i.
the selected light bulb is not fluorescent.
ii.
the selected light bulb is fluorescent given that it is labeled as “long life.”
Are the events fluorescent and long life independent? Explain.
When a person is selected at random from a very large population, the probability that the selected person is righthanded is 0.82. If three people are selected at random, what is the probability that
a.
they are all right-handed?
b.
none of them is right-handed?
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ALGEBRA II
Exit Ticket Sample Solutions
1.
Of the light bulbs available at a store, 𝟒𝟐% are fluorescent, 𝟐𝟑% are labeled as “long life,” and 𝟏𝟐% are fluorescent
and “long life.”
a.
A light bulb will be selected at random from the light bulbs at this store. Rounding your answer to the
nearest thousandth where necessary, find the probability that
i.
the selected light bulb is not fluorescent.
𝟏 − 𝟎. 𝟒𝟐 = 𝟎. 𝟓𝟖
ii.
the selected light bulb is fluorescent given that it is labeled as “long life.”
𝑷(fluorescent given long life) =
b.
𝑷(fluorescent and long life) 𝟎. 𝟏𝟐
=
= 𝟎. 𝟓𝟐𝟐
𝑷(long life)
𝟎. 𝟐𝟑
Are the events fluorescent and long life independent? Explain.
No 𝑷(fluorescent given long life) ≠ 𝑷(fluorescent).
2.
When a person is selected at random from a very large population, the probability that the selected person is righthanded is 𝟎. 𝟖𝟐. If three people are selected at random, what is the probability that
a.
they are all right-handed?
(𝟎. 𝟖𝟐)(𝟎. 𝟖𝟐)(𝟎. 𝟖𝟐) = 𝟎. 𝟓𝟓𝟏
b.
none of them is right-handed?
(𝟎. 𝟏𝟖)(𝟎. 𝟏𝟖)(𝟎. 𝟏𝟖) = 𝟎. 𝟎𝟎𝟔
Problem Set Sample Solutions
1.
When an avocado is selected at random from those delivered to a food store, the probability that it is ripe is 𝟎. 𝟏𝟐,
the probability that it is bruised is 𝟎. 𝟎𝟓𝟒, and the probability that it is ripe and bruised is 𝟎. 𝟎𝟏𝟗.
a.
Rounding your answers to the nearest thousandth where necessary, find the probability that an avocado
randomly selected from those delivered to the store is
i.
not bruised.
𝟏 − 𝟎. 𝟎𝟓𝟒 = 𝟎. 𝟗𝟒𝟔
ii.
ripe given that it is bruised.
𝑷(ripe given bruised) =
iii.
bruised given that it is ripe.
𝑷(bruised given ripe) =
b.
𝑷(ripe and bruised) 𝟎. 𝟎𝟏𝟗
=
= 𝟎. 𝟑𝟓𝟐
𝑷(bruised)
𝟎. 𝟎𝟓𝟒
𝑷(bruised and ripe) 𝟎. 𝟎𝟏𝟗
=
= 𝟎. 𝟏𝟓𝟖
𝑷(ripe)
𝟎. 𝟏𝟐
Which is larger, the probability that a randomly selected avocado is bruised given that it is ripe or the
probability that a randomly selected avocado is bruised? Explain in words what this tells you.
𝑷(bruised given ripe) = 𝟎. 𝟏𝟓𝟖 is greater than 𝑷(bruised) = 𝟎. 𝟎𝟓𝟒. This tells you that ripe avocados are
more likely to be bruised than avocados in general.
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ALGEBRA II
c.
Are the events ripe and bruised independent? Explain.
No, because 𝑷(bruised given ripe) is different from 𝑷(bruised).
2.
Return to the probability information given in Problem 1. Complete the hypothetical 1000 table given below, and
use it to find the probability that a randomly selected avocado is bruised given that it is not ripe. (Round your
answer to the nearest thousandth.)
Bruised
Not bruised
Total
Ripe
𝟏𝟗
𝟏𝟎𝟏
𝟏𝟐𝟎
Not ripe
𝟑𝟓
𝟖𝟒𝟓
𝟖𝟖𝟎
𝑷(bruised given not ripe) =
3.
Total
𝟓𝟒
𝟗𝟒𝟔
𝟏, 𝟎𝟎𝟎
𝟑𝟓
= 𝟎. 𝟎𝟒𝟎
𝟖𝟖𝟎
According to the website www.census.gov, based on the US population in 2010, the probability that a randomly
selected male is 𝟔𝟓 or older is 𝟎. 𝟏𝟏𝟒, and the probability that a randomly selected female is 𝟔𝟓 or older is 𝟎. 𝟏𝟒𝟔.
(In the questions that follow, round your answers to the nearest thousandth.)
a.
If a male is selected at random and a female is selected at random, what is the probability that both people
selected are 𝟔𝟓 or older? (Hint: Use the multiplication rule for independent events.)
(𝟎. 𝟏𝟏𝟒)(𝟎. 𝟏𝟒𝟔) = 𝟎. 𝟎𝟏𝟕
b.
If two males are selected at random, what is the probability that both of them are 𝟔𝟓 or older?
(𝟎. 𝟏𝟏𝟒)(𝟎. 𝟏𝟏𝟒) = 𝟎. 𝟎𝟏𝟑
c.
If two females are selected at random, what is the probability that neither of them is 𝟔𝟓 or older?
If a female is selected at random, the probability that she is not 𝟔𝟓 or older is 𝟏 − 𝟎. 𝟏𝟒𝟔 = 𝟎. 𝟖𝟓𝟒.
So if two females are selected at random, the probability that neither of them is 𝟔𝟓 or older is
(𝟎. 𝟖𝟓𝟒) (𝟎. 𝟖𝟓𝟒) = 𝟎. 𝟕𝟐𝟗.
4.
In a large community, 𝟕𝟐% of the people are adults, 𝟕𝟖% of the people have traveled outside the state, and 𝟏𝟏%
are adults who have not traveled outside the state.
a.
Using a Venn diagram or a hypothetical 1000 table, calculate the probability that a randomly selected person
from the community is an adult and has traveled outside the state.
𝑷(adult and traveled out of state) = 𝟎. 𝟕𝟐 − 𝟎. 𝟏𝟏 = 𝟎. 𝟔𝟏
Lesson 6:
Date:
Probability Rules
2/8/16
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Lesson 6
NYS COMMON CORE MATHEMATICS CURRICULUM
M4
ALGEBRA II
b.
Use the multiplication rule for independent events to decide whether the events is an adult and has traveled
outside the state are independent.
𝑷(adult and traveled out of state) = 𝟎. 𝟔𝟏
𝑷(adult)𝑷(traveled out of state) = (𝟎. 𝟕𝟐)(𝟎. 𝟕𝟖) = 𝟎. 𝟓𝟔𝟏𝟔
Since these two quantities are not equal, the two events are not independent.
5.
In a particular calendar year, 𝟏𝟎% of the registered voters in a small city are called for jury duty. In this city, people
are selected for jury duty at random from all registered voters in the city, and the same individual cannot be called
more than once during the calendar year.
a.
What is the probability that a registered voter is not called for jury duty during a particular year?
𝟎. 𝟗𝟎
b.
What is the probability that a registered voter is called for jury duty two years in a row?
(𝟎. 𝟏𝟎)(𝟎. 𝟏𝟎) = 𝟎. 𝟎𝟏
6.
A survey of registered voters in a city in New York was carried out to assess support for a new school tax. 𝟓𝟏% of
the respondents supported the school tax. Of those with school-age children, 𝟓𝟔% supported the school tax, while
only 𝟒𝟓% of those who did not have school-age children supported the school tax.
a.
If a person who responded to this survey is selected at random, what is the probability that
i.
the person selected supports the school tax?
𝟎. 𝟓𝟏
ii.
the person supports the school tax given that he or she does not have school-age children?
𝟎. 𝟒𝟓
b.
Are the two events has school-age children and supports the school tax independent? Explain how you know
this.
These two events are not independent because the probability of support given no school-age children is not
the same as the probability of support.
c.
Suppose that 𝟑𝟓% of those responding to the survey were over the age of 𝟔𝟓 and that 𝟏𝟎% of those
responding to the survey were both over age 𝟔𝟓 and supported the school tax. What is the probability that a
randomly selected person who responded to this survey supported the school tax given that he or she was
over age 𝟔𝟓?
𝑷(support and over age 𝟔𝟓)
𝑷(over age 𝟔𝟓)
𝟎. 𝟏𝟎
=
𝟎. 𝟑𝟓
≈ 𝟎. 𝟐𝟖𝟔
𝑷(support given over age 𝟔𝟓) =
Lesson 6:
Date:
Probability Rules
2/8/16
© 2014 Common Core, Inc. Some rights reserved. commoncore.org
91
This work is licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
